Calculate the grams of solute needed to prepare.

Question

A. 2.00 L of a 1.50 M NaOH

 

B. 4.00 L of a 0.200 M KCL

 

C. 25.0 mL of a 6.00 M HCI

Answer 1

A.  2.00 L of 1.50 M NaOH

1.50 mole/L x 2.00 L x 39.99 g/mole = 120. g NaOH

B. 4.00 L of 0.200 M KCl

0.200 moles/L x 4.00 L x 74.6 g/mole = 59.7 g KCl

C.  25.0 ml of a 6.00 M HCl

6.00 mole/L x 0.025 L x 36.5 g/mole = 5.48 g HCl

 

All answers are given to 3 significant figures